Starting at home, Luis traveled uphill to the school supply store for 60 minutes at just 6 mph. He then traveled back home along the same path downhill at a speed of 12 mph. What is his average speed for the entire trip from home to the school supply store and back?
Solution: The average speed is not just the average of 6 mph and 12 mph. He traveled for a longer time uphill (since he was going slower), so we can estimate that the average speed is closer to 6 mph than 12 mph. To calculate the average speed, we will make use of the following: $\text{average speed} = \dfrac{{\text{total distance}}}{{\text{total time}}}$ $\text{distance uphill} = \text{distance downhill}$ What was the total distance traveled? ${\begin{align*}\text{total distance} &= \text{distance uphill} + \text{distance downhill}\\ &= 2 \times \text{distance uphill}\end{align*}}$ $\begin{align*}\text{distance uphill} &= \text{speed uphill} \times \text{time uphill} \\\ &= 6\text{ mph} \times 60\text{ minutes}\times\dfrac{1 \text{ hour}}{60 \text{ minutes}}\\ &= 6\text{ miles}\end{align*}$ Substituting to find the total distance: ${\text{total distance} = 12\text{ miles}}$ What was the total time spent traveling? ${\text{total time} = \text{time uphill} + \text{time downhill}}$ $\begin{align*}\text{time downhill} &= \dfrac{\text{distance downhill}}{\text{speed downhill}}\\ &= \dfrac{6\text{ miles}}{12\text{ mph}}\times\dfrac{60 \text{ minutes}}{1 \text{ hour}}\\ &= 30\text{ minutes}\end{align*}$ ${\begin{align*}\text{total time} &= 60\text{ minutes} + 30\text{ minutes}\\ &= 90\text{ minutes}\end{align*}}$ Now that we know both the total distance and total time, we can find the average speed. $\begin{align*}\text{average speed} &= \dfrac{{\text{total distance}}}{{\text{total time}}}\\ &= \dfrac{{12\text{ miles}}}{{90\text{ minutes}}}\times\dfrac{60 \text{ minutes}}{1 \text{ hour}}\\ &= 8\text{ mph}\end{align*}$ The average speed is 8 mph, and which is closer to 6 mph than 12 mph as we expected.